Evaluate the triple integral. $ \int_0^3 \int_0^y \int_{-1}^2 2x + 2z \, dx \, dz \, dy =$
Answer: We can evaluate triple integrals by repeated integration: $ \int_{a_0}^{a_1} \int_{b_0}^{b_1} \int_{c_0}^{c_1} f(x, y, z) \, dx \, dy \, dz = \int_{a_0}^{a_1} \left( \int_{b_0}^{b_1} \left[ \int_{c_0}^{c_1} f(x, y, z) \, dx \right] dy \right) dz$ The first layer: $\begin{aligned} &\int_0^3 \int_0^y \int_{-1}^2 2x + 2z \, dx \, dz \, dy \\ \\ &= \int_0^3 \int_0^y \left[ x^2 + 2xz \right]_{-1}^2 dz \, dy \\ \\ &= \int_0^3 \int_0^y 4 + 4z - 1 + 2z \, dz \, dy \\ \\ &= \int_0^3 \int_0^y 6z + 3 \, dz \, dy \end{aligned}$ The second layer: $\begin{aligned} \int_0^3 \int_0^y 6z + 3 \, dz \, dy &= \int_0^3 \left[ 3z^2 + 3z \right]_0^y \\ \\ &= \int_0^3 3y^2 + 3y \end{aligned}$ The third layer: $\begin{aligned} \int_0^3 3y^2 + 3y &= \left[ y^3 + \dfrac{3y^2}{2} \right]_0^3 \\ \\ &= 27 + \dfrac{27}{2} \\ \\ &= \dfrac{81}{2} \end{aligned}$ In conclusion: $ \int_0^3 \int_0^y \int_{-1}^2 2x + 2z \, dx \, dz \, dy = \dfrac{81}{2}$